Monthly food expenditures for families in Rodriguez Rizal averaged Php6,500 w a standard deviation of Php750. Assuming that the monthly food expenditures are normally distributed. What percentages of these expenditures are between Pho5.900 and Php7,8502

In this case, we need to find the area between X = 5,900 and X =7,850

By using the Z score formula given by:

Z = $\frac{X -\mu}{\sigma}$ 

Given $\mu$ = 6,500 and $\sigma$ = 750

When X = 5,900

Z = $\frac{5,900 -6,500}{750}$  = -0.8

When X = 7,850

Z = $\frac{7,850 -6,500}{750}$  = 1.8

Therefore, we need to find the area between Z = -0.8 and Z = 1.8

That is;

P(-0.8<Z<1.8) = P(Z<1.8) - P(Z<-0.8)

From the standard normal table,

P(Z<1.8) = 0.9641

Also,

P(Z<-0.8) = 0.2119

Therefore,

P(-0.8<Z<1.8) = P(Z<1.8) - P(Z<-0.8) = (0.9641 - 0.2119) = 0.7522

Answer: 0.7522